Question

If A is lies in $Q_3$, and $3\tan A-4=0$ then $3\sin A+4\cos A$ is?

• A. $0$
• B. $\frac{-24}{5}$
• C. $\frac{24}{5}$
• D. $\frac{48}{5}$

Answer 1

The correct option is B $\frac{-24}{5}$

$Q_3\to \tan +ve$

$\sin ,\cos -ve$

$3\tan A-4=0$

$\tan A=\frac{4}{3}$

$\sin A=\frac{-4}{5}$

$\cos A=-\frac{3}{5}$

$3\sin A+4\cos A=3\left(\frac{-4}{5}\right)+4\left(\frac{-3}{5}\right)=\frac{-12-12}{5}=\frac{-24}{5}$